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Question
A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10−8Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?
Solution
Given diameter of the wire,
D = 0.5 × 10−3 m
resistivity of copper, ρ = 1.6 x 10−8 ohm m required resistance, R = 10 ohm
As R = `(ρl)/A, l = (RA)/ρ = (R((piD^2)/4))/ρ = (πRD^2)/(4ρ)` ...`[∵ A = pir^2 = pi(D/2)^2 = (piD^2)/4]`
∴ `l = (3.14 xx 10 xx (0.5 xx 10^-3)^2)/(4 xx 1.6 xx 10^-8)`
m = 122.7 m
Since `R = (ρl)/((πD^2)/4)`
= `(4ρl)/(πD^2), R∝1/D^2`.
When D is doubled, R becomes `1/4` times.
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